BY NATURAL SINES.

AS THE SIDE OPPOSITE THE GIVEN ANGLE IS TO THE NAT. SINE OF THAT ANGLE, SO IS THE OTHER GIVEN SIDE TO THE NAT. SINE OF ITS OPPOSITE ANGLE.

One given side 200, nat. sine of 46° 30', its opposite angle, 0.72537, the other given side 240.

As 200: 0.72537 :: 240; 0.87044-60° 30',

CASE IIL

Fig. 48.
C

Two sides and their contained angle given, to find the other angles and side. Fig. 48.

180

A

36 40

B

240

The solution of this CASE depends on the following PROP

OSITION.

IN EVERY PLANE TRIANGLE, AS THE SUM OF ANY TWO SIDES IS TO THEIR DIFFERENCE, SO IS THE TANGENT OF HALF THE SUM OF THE TWO OPPOSITE ANGLES TO THE TANGENT OF HALF THE DIFFERENCE BETWEEN THEM.

ADD THIS HALF DIFFERENCE TO HALF THE SUM OF THE ANGLES, AND YOU WILL HAVE THE GREATER ANGLE, AND SUBSTRACT THE HALF DIFFERENCE FROM THE HALF SUM, AND YOU WILL HAVE THE LESSER ANGLE.

In the triangle ABC, given the side A B 240, the side A C 180, and the angle at A 36° 40', to find the other angles and side.

The given angle BAC 36° 40', substracted from 180°, leaves 143° 20′, the sum of the other two angles, the half of which

The half sum of the two unknown angles,
The half difference between them,

71° 40'

23 20

Add, gives the greater angle ACB

95 00

48 23

Substract, gives the lesser angle ABC

The side B C may be found by CASE I. or II.

The solution of this CASE depends on the following PROP

OSITION.

IN EVERY PLANE TRIANGLE, AS THE LONGEST SIDE IS TO THE SUM OF THE OTHER TWO SIDES, SO IS THE DIFFERENCE BETWEEN THOSE TWO SIDES TO THE DIFFERENCE BETWEEN THE SEGMENTS OF THE LONGEST SIDE, MADE BY A PERPENDICULAR LET FALL FROM THE ANGLE OPPOSITE THAT SIDE

Half the difference between these segments, added to half the sum of the segments, that is, to half the length of the longest side, will give the greatest segment; and this half difference substracted from the half sum will be the lesser segment. The triangle being thus divided, becomes two right angled triangles, in which the hypothenuse and one leg

In the triangle A B C, given the side A B 105, the side A C

85, and the side B C 50, to find the angles.

Side A C

85

A C

85

Thus the triangle is divided into two right angled triangles, A DC and BDC; in each of which the hypothenuse and one leg are given to find the angles.

To find the angle DCA.

To find the angle DC B.

As hyp. A C, 85 - 1.929419 As hyp. B C, 50

: radius

1.698970

: sine DCA, 61° 56′ 9.945642: sine DCB, 36° 52′ 9.778151

The angle DCA 61° 56′, substracted from 90°, leaves the angle CAD 28° 4′.

The angle DCB 36° 52', substracted from 90°, leaves the angle CBD 53° 16'.

The angle DCA 61° 56′, added to the angle D C B 36° 52′, gives the angle A CB 98° 48'.

This CASE may also be solved according to the following

PROPOSITION.

IN EVERY PLANE TRIANGLE, AS THE PRODUCT OF ANY TWO SIDES CONTAINING A REQUIRED ANGLE IS TO THE PRODUCT

OF HALF THE SUM OF THE THREE SIDES, AND THE DIFFERENCE BETWEEN THAT HALF SUM AND THE SIDE OPPOSITE THE ANGLE REQUIRED, SO IS THE SQUARE OF RADIUS TO THE SQUARE OF THE CO-SINE OF HALF THE ANGLE REQUIRED.

Those who make themselves well acquainted with TRIGONOMETRY, Will find its application easy to many useful purposes, particularly to the mensuration of heights and distances. These are here omitted, because as this work is designed principally to teach the art of common FIELD-SURVEYING, it was thought improper to swell its size, and consequently increase its price, by inserting any thing not particularly connected with that art.

It is recommended to those who design to be surveyors, to study TRIGONOMETRY thoroughly; for though a common field may be measured without an acquaintance with that science, yet many cases will occur in practice, where a knowledge of it will be found very beneficial; particularly in dividing land, and ascertaining the boundaries of old surveys. Indeed no one who is ignorant of TRIGONOMETRY, can be an accomplished surveyor.

SURVEYING is the art of measuring, laying out, and dividing land.

PART I.

MEASURING LAND.

The most common measure for and is the acre; which contains 160 square rods, poles, or perches; or 4 square roods, each containing 40 square rods.

The instrument most in use, for measuring the sides of fields, is GUNTER's chain, which is in length 4 rods, or 66 feet; and is divided into 100 equal parts, called links, each containing 7 inches and 92 hundredths. Consequently, 1 square chain contains 16 square rods, and ten square chains make 1 acre.

In small fields, or where the land is uneven, as is the case with a great part of the land in New England, it is better to use a chain of only two rods in length, as the survey can be more accurately taken.*

PROBLEM I. To reduce two rod chains to four rod

chains.

RULE. If the number of two rod chains be even, take half the number for four rod chains, and annex the links, if any; thus, 16 two rod chains and 37 links, make 8 four rod chains and 37 links.

But if the number of chains be odd, take half the greatest even number for chains, and for the remaining number add 50 to the link: Thus, 17 two rod chains and 42 links makes 8 four rod chains and 92 links.

PROBLEM II. To reduce two rod chains to rods and decimal parts.

As there is no standard of long measure established by the General Government, the standard yard, which is kept in the State Treasurer's office, is adopted as a standard of long measure for Connecticut. Each chain of 2 rods, should be 11, and each of 4 rods, 22 yards, in